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lecture_2_slides
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| lecture_2_slides [2016/01/24 21:45] – created rupert | lecture_2_slides [2017/01/25 10:51] (current) – rupert | ||
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| * Eliminating variables, we get $x=-16+5z$, $y=7-2z$ | * Eliminating variables, we get $x=-16+5z$, $y=7-2z$ | ||
| * The line of intersection consists of the points $(-16+5z, | * The line of intersection consists of the points $(-16+5z, | ||
| + | |||
| + | ==== A detailed look at the last example ==== | ||
| + | * $\begin{array}{ccccccrrr} x& | ||
| + | * Find solutions of this [[system of linear equations|system]] by applying operations | ||
| + | * Aim to end up with a very simple sort of system where we can see the solutions easily. | ||
| + | |||
| + | ==== ==== | ||
| + | * $\begin{array}{ccccccrrr} x& | ||
| + | * Replace equation (2) with $(2)-2\times (1)$: | ||
| + | * $\begin{array}{ccccccrrr} x& | ||
| + | * Now replace equation (1) with $(1)-3\times (2)$ | ||
| + | * $\begin{array}{ccccccrrr} x&&& | ||
| + | |||
| + | ==== ==== | ||
| + | * $\begin{array}{ccccccrrr} x&&& | ||
| + | * can easily rearrange (1) to find $x$ in terms of $z$ | ||
| + | * can easily rearrange (2) to find $y$ in terms of $z$ | ||
| + | * Since $z$ can take any value, write $z=t$ where $t$ is a "free parameter" | ||
| + | * (which means $t$ can be any real number, or $t\in \mathbb{R}$). | ||
| + | * Solution: \begin{align*} x& | ||
| + | |||
| + | ==== ==== | ||
| + | * Solution: \begin{align*} x& | ||
| + | * Can also write this in " | ||
| + | * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix}, | ||
| + | * This is the equation of the line where the two planes described by the original equations intersect. | ||
| + | |||
| + | ==== ==== | ||
| + | * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix}, | ||
| + | * For each value of $t$, we get a different solution (a different point on the line of intersection). | ||
| + | * e.g. take $t=0$ to see that $(-16,7,0)$ is a solution | ||
| + | * take $t=1.5$ to see that $(-16+1.5\times 5, | ||
| + | * etc. | ||
| + | * This works for any value $t\in\mathbb{R}$, | ||
lecture_2_slides.1453671952.txt.gz · Last modified: by rupert